i've forgotten how to integrate fractions!

Hi, A at time t should be:

A(t) = 1000e^{\int_0^t \frac{0.05}{1 + 0.05s}ds} = 1000 + 50t

but, even after searching on google, i can't seem to get their answer. i know this is a simple question but i've just completely forgotten how it's done!

can anyone help? thank you :smile:

Reply 1

nuodai

In general,

\displaystyle \int^t \dfrac{1}{a+bx}\, dx = \dfrac{1}{b} \log (a+bt)

; you can see this by making the substitution

y=a+bx

Reply 2

Ewan

Do you remember what to do when the top is the differential of the bottom? if you want make a substitution x = 1+0.05s, it might make it more obvious.

Reply 3

Farhan.Hanif93

Original post by troola

Hi, A at time t should be:

A(t) = 1000e^{\int_0^t \frac{0.05}{1 + 0.05s}ds} = 1000 + 50t

but, even after searching on google, i can't seem to get their answer. i know this is a simple question but i've just completely forgotten how it's done!

can anyone help? thank you :smile:

It might be helpful to notice that

\displaystyle\int_0^t \dfrac{0.05}{1 + 0.05s}ds \equiv \displaystyle\int_0^t \dfrac{\frac{d}{ds}[1+0.05s]}{1 + 0.05s}ds

.

What do you know about integrals of the form

\displaystyle\int \dfrac{f'(x)}{f(x)}dx

?

EDIT: Too slow.

Reply 4

troola

i used to have a list of rules for different kinds of integrals but i've lost them, and since i haven't done this kind of maths for so long i've completely forgotten it!

following what nuodai said i have

\frac{0.05}{0.05}log(1+0.05t)

which i just know is wrong! hmph :frown:

Reply 5

troola

Original post by Farhan.Hanif93

It might be helpful to notice that

\displaystyle\int_0^t \dfrac{0.05}{1 + 0.05s}ds \equiv \displaystyle\int_0^t \dfrac{\frac{d}{ds}[1+0.05s]}{1 + 0.05s}ds

.

What do you know about integrals of the form

\displaystyle\int \dfrac{f'(x)}{f(x)}dx

?

EDIT: Too slow.

no i didn't know about this, what would be the rule when the top is the differential of the bottom? it seems there are a lot of different rules for a lot of different integrals! :s-smilie:

Reply 6

nuodai

Original post by troola

\frac{0.05}{0.05}log(1+0.05t)

which i just know is wrong! hmph :frown:

What makes you think it's wrong? Put that in and simplify it.

Reply 7

Farhan.Hanif93

Original post by troola

\frac{0.05}{0.05}log(1+0.05t)

which i just know is wrong! hmph :frown:

For the integration part, that's right. If you don't believe me, differentiate it back and see what you get.

Reply 8

gozatron

Original post by troola

Hi, A at time t should be:

A(t) = 1000e^{\int_0^t \frac{0.05}{1 + 0.05s}ds} = 1000 + 50t

but, even after searching on google, i can't seem to get their answer. i know this is a simple question but i've just completely forgotten how it's done!

can anyone help? thank you :smile:

Just from another vieewpoint, it may be helpful for you to think about what you get when you differentiate a log function.

Reply 9

troola

oh ok my bad!

so would i end up with

1000e^{log(1+0.05t)}

or am i going completely off track?

Reply 10

Farhan.Hanif93

Original post by troola

no i didn't know about this, what would be the rule when the top is the differential of the bottom? it seems there are a lot of different rules for a lot of different integrals! :s-smilie:

If you consider the following:

\dfrac{d}{dx}[\log f(x)]

Where this is the natural log of f(x); if you let

y=\log f(x)

and

u=f(x)

, we have

y=\log (u)

. Therefore

\dfrac{du}{dx} = f'(x)

and

\dfrac{dy}{du}=\dfrac{1}{u}= \dfrac{1}{f(x)}

.

By the chain rule,

\dfrac{d}{dx} [\log f(x)] = \dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx} = f'(x) \times \dfrac {1}{f(x)} = \dfrac{f'(x)}{f(x)}

Then, if you remember that integration is the reverse operation for differentiation, notice that

\dfrac{d}{dx}[\log f(x)] = \dfrac{f'(x)}{f(x)} \implies \displaystyle\int \dfrac{f'(x)}{f(x)} dx = \log f(x) + C

(edited 13 years ago)

Reply 11

Farhan.Hanif93

Original post by troola

oh ok my bad!

so would i end up with

1000e^{log(1+0.05t)}

or am i going completely off track?

That's right, and since

e^{\ln K}=...?

Reply 12

R0BMAN

Original post by troola

no i didn't know about this, what would be the rule when the top is the differential of the bottom? it seems there are a lot of different rules for a lot of different integrals! :s-smilie:

That was the first thing I learnt when integrating fractions. Always look to see if the top is the differential of the bottom. If you differentiate say, log(2x+1), you differentiate the bracket then do that over the bracket, if you get me. f'(x)/f(x).

So when you're integrating

you'd get 1000e^[log(1+0.05s)] - with limits of t and 0 - but log1 is just zero, so it would be log(1+0.05t) - e^ log anything just equals the bracket, so it would be 1000(1+0.05t) which is 1000+50t.

Make sense?

Reply 13

troola